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4 years ago

13

A lock has 60 digits, and the combination involves turning right to the first number, turning left to the second number, and tur

ning right to the third number. How many possible combinations are there?

1 answer:

Oduvanchick [21]4 years ago

5 0

Answer: 180

Explanation: For the first number of the combination you have 60 possible options, because you need three numbers for the combo, times it by three.

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The admission fee at a movie theater is $5 for children and $9 for adults. If 3200 people go to the movies and $24000 is collect

Usimov [2.4K]

For this problem, let x be the number of children and y for adults. Formulate the equations: 1st equation, x + y = 3,200 and 2nd equation 5x + 9y = 24,000. Re-arrange 1st equation into x = 3200 - y. Then, substitute into 2nd equation, 5(3,200-y) + 9y = 24,000. Then, solve for y. The 16,000 - 5y + 9y = 24000. Final answer is, y = 2000 adults went to watch the movie.

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4 years ago

Please helpppp meee !!!!

Marta_Voda [28]

Answer:

y=-9x+193

Step-by-step explanation:

Its decreasing by 9, and 175+(9x2) is 193

6 0

3 years ago

Read 2 more answers

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

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3 years ago

Use the greatest common factor and the distributive property to write and an equivalent expression: 20g+24h

n200080 [17]

4(5g+6h)

You xan remove four from each number, but no letters since they dont have any of the same.

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3 years ago

Explain why it is important to solve for the variable first in order to find the measure any angle mentioned in the problem . Ye

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In the field of astronomy, the ability to measure angles accurately and precisely enables us to calculate the position and relative movement of the stars and galaxies in relation to each other, to determine how far distant they are from us, and even to estimate their relative size.

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3 years ago